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MECHANICS 
OF FLUIDS 



MECHANICS OF FLUIDS 



Brief Notes for Engineering Students 



By JAMES E. BOYD 



The Ohio State University, Columbus, Ohio 

Published by the Author 

1917 



PRESS OF THE F. J. HEER PRINTING CO. 
COLUMRUS, OHIO 

1917 



^ 



\k° 



^ 



COPYRIGHT, 1917 

BY 
JAMES E. BOYD 



/ 

©C!,A4()7059 

MAY 21 1917 

11- J^?f>4^ 



THE MECHANICS OF FLUIDS 



i. Definitions. — Fluids differ from solids in the fact 
that they offer no resistance to change of form, provided that 
changes take place slowly. A solid may be supported by a single 
force or a system of parallel forces. To keep a fluid in equili- 
brium requires forces in more than one direction. 

The term fluid embraces both liquids and gases. A gas has 
no fixed volume, but expands so as to entirely fill the vessel 
which contains it, exerting a pressure which varies inversely as 
the volume (provided the temperature is kept constant). A 
given mass of liquid has a fixed volume, which is changed very 
little by great changes of pressure. A volume of liquid may have 
a free surface at which it exerts no pressure ; a gas must be con- 
fined on all sides. 

Hydraulics is that branch of engineering science which 
treats of the motion of liquids. Hydrostatics treats of liquids 
at rest. The most important liquid is water. Pneumatics treats 
of the mechanics of gases. 

2. Fluid Pressure. — Since fluids offer no resistance to 
change of form, provided that change takes place slowly, it is 
evident that a mass of fluid can exert no shearing stress. The 
forces at the surface of a fluid must, therefore, be normal to the 
surface. A fluid can exert very little tension, so that the only 
force of any considerable magnitude at any surface is a normal 
pressure. This applies to the surface of contact of a fluid and 
a solid, the surface of contact of two fluids, or to any surface 
which may be imagined as separating two portions of the same 
fluid. 

At any point in a fluid the pressure per unit area is the same 
in all directions. 

Fig. i represents a small portion of fluid in the form of a 
triangular prism of length dz. Each base of the prism is a right 
triangle of which the vertical leg is dy, the horizontal leg is dx, 
and the hypotenuse is ds. The hypotenuse makes an angle 6 




Fig. 1. 



the inclined surface, 



with the horizontal. Let p be the 
pressure per unit area on the upper 
surface. The total pressure on this 
surface is p dx dz. This vertical 
pressure must be balanced by the 
vertical component of the normal 
pressure on the inclined surface of 
area dz ds, if the prism is so small 
that its own weight is negligible. The 
two triangular surfaces and the 
rectangular surface dy dz, being verti- 
cal, can exert no vertical force. If 
N represent the total pressure and p' 
represent the unit pressure normal to 



N = p' dsdz. 

The vertical component of N is N cos 6 ; 

p dx dz = p' ds dz cos 0. 
Since dx = ds cos 0, 

p = p\ 



(i) 

(2) 

(3) 



By resolving horizontally parallel to dx, the unit pressure 
on the vertical rectangle dy dz may likewise be shown to be equal 
to p f and, consequently, equal to p. Since may be any angle 
whatever, and since the axes may be in any direction, the propo- 
sition is proved. 

In a stationary liquid the unit pressure at all points in a 
horizontal plane is constant. 

Suppose a particle is moved slowly from one point to an- 
other in the same horizontal plane. No work ,is done on the 
particle by gravity, since its distance from the center of the 
earth is not changed. If at any point the horizontal pressure 
on the particle is not the same on all sides, this difference of 
pressure would give it kinetic energy. It is evident that the 
liquid will not remain stationary if there is any difference in the 
unit pressure at points at the same level. 

The total pressure on a horizontal surface in a liquid is given 
by the expression, 

Total pressure = zvhA, (4) 




where w is the weight of the liquid per unit volume, h is the 
vertical depth of the surface considered below the free surface 
of the liquid, and A is the area of the surface on which the pres- 
sure acts. 

Consider a vertical column of liquid of base A and height 
h. Fig. 2. Its volume is hA, and its mass is whA. Since a liquid 
can exert no tangential force (except the 
very small force of capillarity), none of 
its weight can be supported by the verti- 
cal walls of the vessel, and, consequently, 
the entire weight whA rests on the sur- 
face A. This is true also if the liquid 
is enclosed laterally by other liquid in- 
stead of the solid vessel. 

The column of liquid may not be verti- 
cal and may not be of the same size 
throughout, but since the pressure per 
unit area is the same at all points at any 
one level, formula (4) is still true. The height h is the vertical 
distance of the surface A below the plane of the liquid surface. 
A cubic foot O'f water at the maximum density (39.2 Fahr.) 
weighs 62.428 pounds. The density of mercury is 13.596 that 
of water. For ordinary purposes, the weight of one cubic foot 
of water is taken as 62.5 pounds and the specific gravity of 
mercury is 13.6. A column of water one foot long and one 
square inch cross-section weighs 0.434 pound. A column of 
water of one square inch cross-section and 2.3 feet high weighs 
one pound. One inch of mercury is equivalent to 1.1333 feet 
of water, or 30 inches of mercury equal 34 feet of water. 

The density of air and other gases is so small that, for 
problems involving liquid pressure, the weight of several feet of 
gas may be neglected without appreciable error. 



Fig. 2. 



PROBLEMS 



1. A cylindrical tank, 2 feet in diameter, is filled with water to a 
depth of 12 feet. Find the total pressure on the bottom. 

Ans. 23,562 pounds. 

2. What is the height of a column of water which exerts a pressure 
of 50 pounds per square inch? 

Ans. 115 feet. 



3. A cylindrical tank, with axis vertical, is 2 feet in diameter and 
4 feet high. A vertical pipe, 1 foot inside diameter, is attached to the 
top of the cylinder and filled with water to a height of 8 feet above the 
top of the cylinder. Find the total pressure on the bottom of the cylin- 
der. Find the weight of water in the cylinder, and the weight of water 
in the pipe. 

Ans. 23,562 pounds; 7854 pounds; 3927 pounds. 

4. In Problem 3, find the total vertical pressure pushing upward 
against the top of the tank. Show that the sum of this pressure and 
the total weight of the water in the tank and pipe is equal to the 
pressure on the bottom. 

3. Measurement of Pressure. — The pressure of a liquid 
or gas is expressed by engineers in feet of water, in inches of 
mercury, or in pounds per square inch. High gas pressures are 
measured in atmospheres. 

Fig. 3 shows a cylindrical vessel filled with water. A verti- 
cal tube, open at the top, is connected to show the pressure. The 
water rises in this tube to a height h above the axis of the 
cylinder. The pressure of a liquid must be taken with reference 
to some definite level. In Fig. 3, this level is the horizontal plane 
through the axis. 

At the right of the cylinder is a U-tube containing mercury. 
The pressure of the water in the cylinder, acting through the air 
in the inverted U part of the tube, has displaced the mercury 
sc that it stands -with its surface in the right leg of the tube at 
a height h 1 above its surface in the left leg. If h 1 is expressed 
in inches, the pressure in the cylinder is h\ inches of mercury. 

In using a gage of any kind to measure fluid pressure, the 
fluid in the connections must be taken into account. If the fluid 
is a gas, its weight may ordinarily be neglected. In the case of 
the U-tube of Fig. 3, the connection is made to the cylinder at 
the plane of reference, and the water in the connection does not 
rise above this plane. Since the weight of the intervening air 
is negligible, the difference in the height of the two mercury 
columns measures the desired .pressure. If there were no air 
in the tube and water filled the connection from the cylinder to 
the mercury at C, the mercury column would then measure the 
pressure with reference to the horizontal plane through C. In 
that case the mercury height h x would be equivalent to a water 
column h -\- h 2 . 



Another type of mercury gage is shown at A. The connect- 
ing tube is entirely filled with water, and the mercury column h z 
measures the difference between the water columns h and h±. 

Steam gages are also used to measure liquid pressure. They 
generally read in pounds per square inch, but are sometimes 



-V&ter 




graduated in feet of water. In using a steam gage to measure 
liquid pressure, care must be taken to have the connections en- 
tirely filled with water, and correction must be made for the 
height of the gage above the plane of reference. 



PROBLEMS 

1. The pressure in a water pipe, measured by a gage which is 5 
feet above the axis of the pipe, is 40 pounds per square inch. What is 
the pressure at the axis of the pipe in feet of water? 

Ans. 97 feet. 

2. A rectangular tank is 4 feet long and 3 feet wide. A mercury 
gage in the form of a U-tube is attached to the tank. The mercury on 
the side nearest the tank reads 12 inches, and on the other side it reads 
32 inches. The zero of the scale is 1 foot above the bottom of the 
tank. Find the total pressure on the bottom in pounds. 

Ans. 17,000 pounds. 

4. Vacuum. — In ordinary gages the pressure is referred 
to atmospheric pressure as the zero. The tubes of Fig. 3 are 







all open to the air, so that they have the atmospheric pressure 
on one side. The height of the liquid column, in each case, 
measures the difference between the pressure in the cylinder and 
the atmospheric pressure. Steam gages, also, measure the ex- 
cess of the pressure above that of the atmosphere. 



It frequently happens that the pressure in a vessel is below 
that of the atmosphere. This pressure is then called a vacuum. 
Fig. 4 shows a cylinder which is connected to a tube extending 
down into an open vessel of liquid. The pressure in the cylinder 
is lower than the atmospheric pressure, so that the liquid is 
forced up in the tube to height h above the free surface in the 
vessel. This column of height h measures the vacuum in the 
cylinder. If the liquid is water, and h is in feet, the vacuum is 
h feet of water. 

A vacuum is measured in feet of water, in inches of mer- 
cury, or in pounds per square inch. 

The difference between the vacuum and the atmospheric 
pressure is the absolute pressure. With a vacuum of 24 inches 
of mercury and a barometer of 29 inches, the absolute pressure 
is 5 inches of mercury. 

PROBLEMS 

1. A tube is connected to a tank and dips into a vessel of mercury. 
The mercury rises to a height of 24 inches in the tube. The barometer 
reads 29.2 inches. Find the vacuum and the absolute pressure in feet of 
water and in pounds per square inch. 

Ans. 27.20 feet; 11.80 pounds per square inch; 
5.89 feet ; 2.56 pounds per square inch. 

2. A bucket 1 foot in diameter is filled with water and, inverted in 
a tank of water and lifted until the bottom is 2 feet above the surface 
of the water in the tank. What is the vacuum at the bottom of the 
inverted bucket, and what is the force required to lift it? 

5. Liquid Pressure on a Plane Surface. — The pressure 
of a liquid on a plane surface is equal to the weight of a column 
of the liquid whose base is the area of the surface and whose 
height is the distance of the center of gravity of this area below 
the surface of the liquid. 

In Fig. 5, G E F represents the horizontal surface of the 
liquid. M N is a plane surface subjected to the liquid pressure. 
It is required to find the total pressure on M N. 

The pressure on a horizontal element of area dA, at a depth 
y below the surface of the liquid, is zvydA. Since liquid pres- 
sure is the same in all directions, the -total pressure on an inclined 
element dA is likewise w y dA, provided all of the element is at 



10 







Fig. 5. 

the same distance y below the liquid surface. The element dA 
has a finite length in the horizontal direction and an infinitesimal 
width at right angles thereto, so that the depth at all points dif- 
fers an infinitesimal amount from the depth y. 

Pressure on dA = iv y dA. ( I ) 

Total pressure on area = wfydA. (2) 

In finding the center of gravity of an area, 

CydA 



hence, C y dA = y A, 

w jy dA = w y A, 

which proves the proposition. 



(3) 



PROBLEMS 

1. A rectangular tank, 5 feet long, 4 feet wide, and 6 feet high, 

is filled with water. Find the total pressure on the bottom, on one side, 
and on one end. 

Ans. 7500, 5625, 450O pounds. 

2. In Problem 1, find the total pressure on the lower 2 feet of one 
side. 

Ans. 3125 pounds. 



1 I 



3. A triangular trough is 5 feet long, 6 feet wide at the top, and 
4 feet deep. Find the total pressure on one end and on one side when 
the trough is filled with water. 

Ans. 1000 pounds on one end ; 5000 pounds on one side. 

4. A cylindrical tank, with axis horizontal, is 4 feet in diameter. 
It is connected to a 2-inch pipe, and tank and pipe are filled with water 
which stands at a height of 12 feet above the axis of the tank. Find 
the total pressure on one end. Ans. 9428 pounds. 

5. A semi-circular tank, with axis horizontal, is 8 feet in diameter. 
Find the total pressure on one end when it is filled with water. 

Ans. 5333 pounds. 

6. Center of Pressure. — The center of pressure on an 
area is the point of application of the resultant pressure. The 
point of application of the resultant of a set of parallel forces 
is obtained by dividing the sum of the moments of all the forces 
with respect to an axis by the sum of the forces. The quotient 
is the distance of the point of application of the resultant from 
the given axis. This is the method used in finding the center 
of gravity. 

In Fig. 6, take moments with respect to the line B D , which 
is the intersection of the plane M N with the surface of the 




Fig. 6. 

liquid. The element dA is parallel to B D and its moment arm 
is y'. The vertical depth of dA below the surface of the liquid 
is y' cos 6, where 6 is the angle which the plane M N makes with 
the vertical. 






12 

Pressure on dA — w y' cos OdA. (i) 

Moment about B D = wy' cos OdA X / = w cos 6y' 2 dA. (2) 
Total moment = w cos 6 C y' 2 dA = w cos 1, (3) 

where / is the moment of inertia of the plane area M N with 
respect to axis B D. 

The total pressure on M N is wy cos A, where y is the 
distance of the center of gravity of the area M N from the line 
BD. 

w cos 01 I 

y c = = , (4) 

w y cos A y A 

where y c is the distance of the center of pressure from the line 
BD. 

The center of pressure of a vertical rectangular area, sub- 

2 h 

jected to liquid pressure on all of one side, is from the 

o 
from the top, where h is the total height. The moment of 

b h s h 

inertia is ; the area is b h, and y is - — . 

3 2 

b h* b h 2 2 h 
y c = ; = . 



323 

In the case of a vertical circular area of radius a, 

5tt a 4 5a 
/ = ; y —a ; and A = tto 2 ; so that y c = 



PROBLEMS 

1. A vertical rectangular gate, 8 feet wide and 9 feet high, is sub- 
jected to water pressure on one side. The gate is hinged at the bottom 
and held by a chain at the top. Find the tension on the chain by taking 
moments around the bottom. 

20,250X3 

Ans. P = " ()750 pounds. 

9 

2. Find the tension on the chain in Problem 1 if the gate is sub- 
jected to a pressure of 9 feet of water on one side and 6 feet of water 
on the other. 

Ans. 4750 pounds. 



13 

3. Find the center of pressure on a vertical rectangular gate, 12 
feet in height, due to water which rises 4 feet above the top of the gate. 

Ans. y c =1116 -v- 108 = 10.33 feet below the surface of the water. 

4. Find the center of pressure on a vertical rectangular gate, 10 
feet high, due to water which rises 4 feet above the top of the gate. 

Ans. 4.10 feet above the bottom. 

Equation (4) may be written, 

/ k 2 A k 2 

3'c = = = , (5) 

y A y A y 

where k is the radius of gyration of the area under pressure with 
respect to the intersection of its plane with the surface of the 
liquid, at a distance y from its center of gravity. The formula 
for the transfer of the moment of inertia of an area is 

/ = I + A d\ 

k 2 = k 2 + d 2 = k 2 + f, 

Vc = -|^ + ^ (6) 

Equation (6) shows that the center of pressure is always 

k 2 
below the center of gravity and is at a distance — ° — therefrom, 

where k is the radius of gyration of the surface under pressure, 
with respect to a horizontal line in its plane through its center of 
gravity. 

PROBLEMS 

5. Find the center of pressure on the end of the cylinder of Prob- 
lem 4 of Article 5 by means of equation (6). 

Ans. i foot below the axis of the cylinder. 

6. Solve Problem 5 if the water rose 8 feet above the axis of the 
cylinder. 

Ans. 1 inch below the axis. 

7. In Problem 3 of Article 5, find the' center of pressure on one 
end. 

Ans. 2 feet from the top. 

7. Pressure on a Curved Surface. — Equation (3) of 
Article 5 also gives the total pressure on a curved surface. As 
the pressure is always normal to the surface, this total pressure 
is not the resultant pressure in any one direction. 



14 



Fig. 7, I, shows a portion of fluid enclosed by a plane and 
a curved surface. The resultant pressure on the curved surface 
in the direction normal to the plane surface is P. The com- 
ponents of the pressure on the curved surface, which are parallel 
to the plane surface, may be divided into two groups, the result- 
ants of which are equal and opposite. 





Fig. 7. 



Fig- 7, 



II, represents the equilibrium of the fluid portion 
between the two surfaces. The pressure of the plane A B against 
the fluid, which is equal to the pressure of the fluid against A B, 
is represented by N. The component of the total pressure of 
the curved surface parallel to N is represented by P. The 
weight of the portion of fluid between the surfaces is W. Re- 
solving perpendicular to A B, 

N+WcosO = P, (i) 

where 6 is the angle which N makes with the vertical, or the 
angle which the plane A B makes with the horizontal. If the 
fluid is a gas, its weight is negligible, and equation (i) becomes 

N = P. (2) 

If p is the unit pressure on the plane surface A B, and A is the 
area, 

N — pA. (3) 

The pressure of a gas in one direction on a curved surface 
is equal to its total pressure on the area obtained by projecting 
the curved surface upon a plane which is normal to the given 
direction. 

The pressure of a liquid is found in the same way, with 
the addition of the component of the weight of the portion of 
the liquid between the curved and the plane surfaces. 



i5 

PROBLEMS 

1. A hollow cylinder of length /, and inside diameter D, is sub- 
jected to a steam pressure of p pounds per square inch. Find the total 
force tending to split the cylinder longitudinally. 

Ans. Total force = p I D. 

2. The cylinder of Problem 1 is made of steel plates of thickness 
t. Find the circumferential unit stress which resists longitudinal rupture. 

P D 

Ans. s = . 



2 t 

3. A boiler shell, 5 feet in diameter, is made of i-inch steel 
plates. If the allowable unit tensile stress is 9000 pounds per square 
inch, find the allowable internal pressure. 

Ans. 150 pounds per square inch. 

4. A hollow cylinder, of inside diameter K D, is subjected to an 
internal pressure of p pounds per square inch. Find the total force on 
one end of the cylinder in trfe direction of its length. 

5. In Problem 4, find approximately the axial unit stress if the 
thickness is t. 

Ans. j : 

6. A cylinder, with axis vertical, is 4 feet in diameter and 5 feet 
long. The top is connected to a cone 6 feet high inside, which is bolted 
to the cylinder. At the bottom a hollow hemisphere of 2-foot radius is 
fastened to the cylinder. A pipe is connected to one side. The cylinder 
(including the hollow cone and hemisphere) and pipe are filled with 
water. Find the total upward pressure on the cone and the total down- 
ward pressure on the hemisphere when the water in the pipe stands 10 
feet above the top of the cylinder. 

Ans. Upward pressure on cone = 7854 — 1571 = 6283 pounds ; 
downward pressure on hemisphere = 11,781 -4- 1047 = 
12,828 pounds. 

7. The hemispherical end of the cylinder of Problem 6 weighs 600 
pounds. It is fastened to the cylinder by eight bolts. Find the diameter 
of the bolts if the allowable unit stress is 6000 pounds per square inch. 

8. Volume Change Due to Temperature and Pressure, — 

The expansion of a liquid with change of temperature is much 
greater than the expansion of a solid. Water is peculiar in the 
fact that it contracts when the temperature rises from o° to 4 
Centigrade (32 to 39. 2° Fahr.), and then expands with further 
increase of temperature. Table I gives the relative volume and 
weight per cubic foot for pure water. 



i6 

TABLE I 
Relative Volume and Density of Water. 






Temperature in 
degrees Fahr. 


Relative volume 


Weight per cubic 
foot, in pounds 


32 


1.00013 


62.42 




39.2 


1.00000 


62.428 




50 I 


1.00027 


62.41 




60 


1,00096 


62.37 




70 


1.00201 


62.30 




80 


1.00338 


62.22 




90 


1.00504 


! 62.12 




100 


1.00698 


62.00 




150 


1.02011 


61.21 




212 


1.04343 


59.84 








The U. S. liquid gallon contains 231 cubic inches. The 
British Imperial gallon contains 277.41 cubic inches, which is 
10.0221 pounds of water at the maximum density. For ordinary 
calculation, a cubic foot may be taken as weighing 62.5 pounds 
and equal to 6.25 British Imperial gallons. 

Liquids are frequently regarded as incompressible. Com- 
pared with a gas, this is true, but compared with most solids, 
liquids are very compressible. The modulus of volume elasticity 
of water is about 300,000 pounds per square inch,* which means 
that one pound per square inch will diminish the volume of a 
mass of water one part in 300,000, or that 3,000 pounds per 
square inch will diminish the volume one percent. The corres- 
ponding modulus for steel is about 20,000,000 pounds, per square 
inch, or nearly seventy times as great. 

PROBLEMS 

1. A hot water heating system holds 10 cubic feet of water and 
is connected to a cylindrical expansion tank which is 10 inches inside 
diameter. How much will the water rise in the tank when the temper- 
ature of the water rises from 40° to 212° Fahr? 

Ans. 9.7 inches. 



* See Landolt and Bornstein's Tables, page 31, or Smithsonian 
Physical Tables, page 82, for the modulus of volume elasticity. 



17 



2. A vessel containing 6 gallons is filled with water at 39.2° Fahr. 
The temperature is then raised to 150° Fahr. How much water will 
flow out, if the vessel does not expand? 

Ans. 27.8 cubic inches. 

3. If the material of the vessel of Problem 2 has a coefficient of 
linear expansion of 0.000007 per degree Fahrenheit, how much water 
will flow out? 



9. Velocity of Flow from an Orifice. — Fig. 8 shows an 
orifice in a vessel of liquid. The center of the orifice is at a 

distance h below the liquid surface. 
If w pounds of the liquid moves 
downward from the surface of the 
liquid to the orifice, the work 
of gravity on it is w h foot pounds. 
If there is no loss by friction, the 
mass of w pounds must have w h 
foot-pounds of kinetic energy when 
it issues from the orifice. If V is 
the velocity of the jet, the kinetic 
energy of w pounds is 
wV 2 




*g 



Fig. 8. 



foot-pounds. 

wV 2 
ft = 

29 
V 2 = 2gh, 



7c' 



V = V 2 g h. 



Co 



Formula I. 



Formula I gives the velocity with which liquid flows from 
an orifice when there is no loss of energy. It is the velocity 
which a freely falling body acquires in falling a distance h. This 
velocity is frequently called the theoretical velocity. In the case 
of a liquid of little viscosity, such as water, the theoretical 
velocity is the actual velocity. With viscous liquids, such as 
heavy oils, the actual velocity is considerably less than the theo- 
retical velocity. Formula I is based on the assumption that the 
atmospheric pressure on the orifice and the free surface of the 
liquid is the same. 



i8 



PROBLEMS 

1. Water flows from an orifice under a head of 32 feet. Ii g = 
32.2, what is the theoretical velocity? 

Ans. 45.4 feet per second. 

2. An orifice in a tank is 20 feet below the surface of the water 
in the tank. The upper surface of the water is subjected to an air 
pressure of 25 pounds per square inch above the atmosphere, while the 
orifice opens into the free air. Find the velocity of flow from the 
orifice. 

Ans. 70.6 feet per second. 

3. A large pipe reaches down into a pond of water which is 
under atmospheric pressure. At a point 12 feet above the surface of 
the pond, the pipe ends with an orifice which opens into a tank in 
which there is a vacuum of 10 pounds per square inch. With what 
velocity will the water flow from the pipe into the tank? 

Ans. 26.6 feet per second. 

io. Coefficients of Velocity, Contraction, and Discharge. 

— - The ratio of the actual velocity in an orifice or nozzle to the 
theoretical velocity is called the coefficient of velocity. It is 
represented by K y . 

Actual velocity 

K, = — — . ( i ) 

V 2 g h 




Fig. !». 



When an orifice is connected to a pipe 
or tank of considerably larger diameter, 
the momentum of the liquid flowing in 
from all sides tends to carry it beyond 
the edge of the orifice. Fig. 9 shows 
the direction of flow at different parts 
of the stream. In the plane of the ori- 
fice the liquid near the edges makes a 
considerable angle with the normal to 
the plane of the orifice. Nearer the 
center the angle is less. At section B, 
at some distance outside the plane of the 
orifice, the liquid flows in the same di- 
rection at all points of the section, and 
t 1 e area of the cross-section of the jet 
is smaller than the area of the orifice. 



19 

The position where the jet ceases to converge is called the con- 
tracted vein. The ratio of the area of the cross-section of the 
jet at the contracted vein to the area of the orifice is called the 
coefficient of contraction. It is represented by K c . In a circular 
orifice the coefficient of contraction is the ratio of the square of 
the diameter of the jet to the square of the diameter of the ori- 
fice. 

The quantity of liquid which flows past any section of a 
pipe or jet in unit time is the product of the area of the section 
multiplied by the velocity of the liquid. 

Q=AV, (2) 

where is the quantity in unit time and A is the area of the 
section. At the contracted vein V = K v \/ 2 g h, and area of 
section = K C A, where A is the area of the orifice. 



Q = K x K c Ay 2gh. (3) 



A \/ 2 g h gives the quantity which would flow through the 
orifice in unit time if there were no contraction and the actual 
velocity were the same as the theoretical velocity. This is multi- 
plied by the product K V K C to get the actual quantity discharged. 
This product may be replaced by a single letter K&, and the term 
called the coefficient of discharge. 

Q 

K d = ; 



A V 2 gh 

it is the ratio of the actual quantity discharged to the quantity 
which would flow through a section of area A, if the velocity 
were the theoretical velocity. 

Q = K d A ^f~2gh. Formula II. 

PROBLEMS 

1. The diameter of an orifice is 2 inches and the diameter of the 
jet at the contracted vein is 1.6 inches. Find the coefficient of con- 
traction. 

Ans. K c = 0.64. 

2. Water flows from a nozzle under a head of 20 feet. The 
average velocity of the jet is 34 feet per second. Find the coefficient 
of velocity. 

Ans. K r = 0.95. 



20 



3. If the coefficient of velocity is 0.98 and the diameter of the jet 
is 0.9 inch when the diameter of the orifice is 1 inch, what is the co- 
efficient of discharge? 

Ans. K d = 0.794. ' 

4. The discharge from a 4-inch circular nozzle under a head of 
12 feet is 2 cubic feet per second. Find the coefficient of discharge. 

Ans. K d = 0.824. 

ii. Orifice in a Thin Plate. — Fig. io shows a so-called 
orifice in a thin plate. As a thin wall will not sustain the pres- 
sure of the liquid behind it, the orifice is actually made in a rela- 
tively thick plate, but with the 
outer portion conical, and the 
inner portion cylindrical for a 
small thickness. The liquid does 
not touch the conical surface, so 
that the orifice is equivalent to 
an orifice in a plate of thickness 
equal to the length of the cylin- 
drical portion. Usually the liquid 
touches the cylindrical surface 
at the inner edge only. If care 
is taken to prevent the liquid 
from wetting the cylindrical sur- 
face at the beginning, the entire 
orifice may be cylindrical and 
still act as an orifice in a thin 
plate ; but if the surface of such 
a cylinder be once wet, the capil- 
larity will cause the liquid to 
adhere to it and the orifice will 
behave like a short pipe (Article 24). 

In an orifice in a thin plate the coefficient of velocity is 
practically unity. For this reason Professor S. W. Robinson 
called these orifices frictionless orifices. (The coefficient of 
velocity is given by many authors to be about 0.98. This figure 
is based on the range of the jet in air, and is not correct.*) 

The contracted vein is at a distance from the plane of the 
back of the orifice a little greater than the radius of the orifice. 





* Some experiments with Frictionless Orifices, by Judd and King, 
Engineering News, Sept. 27, 1906, page 326. 



21 



The diameter of the jet at the contracted vein is a little less than 
0.8 the diameter of the orifice. The coefficient of contraction 
is about 0.62 and the coefficient of discharge, since K v equals 
unity, is the same. 

PROBLEMS 

1. How much water will flow in one second from a 6-inch orifice 
in a thin plate under a head of 24 feet of water if K d = 0.62 and 
g = 32.16? 

Ans. 7.71 cubic feet. 

2. A circular orifice 1.9975 inches in diameter discharged 50.2 
cubic feet of water in 211.8 seconds under a head of 4.978 feet, measured 
from the center of the orifice. Find the coefficient of discharge. 

Ans. K d = 0.609. 

3. The orifice of Problem 2 discharged 50.7 cubic feet in 211.3 
seconds under a head of 4.996 feet. Find the coefficient of discharge. 

Ans. i^ d = 0.615. 

4. The average diameter of the jet from the orifice of Problem 
2, at a distance of 1.10 inches from the back of the plate, was 1.5678 
inches. Find the coefficient of contraction. 

Ans. K c = 0.616. 

12. Orifice with Contraction Suppressed. — Orifices in 
thin plates are much used to measure the quantity of flow. In 
order that the coefficient of discharge may be known, it is neces- 
sary that the back of the plate shall be a plane surface many 
times larger than the orifice. In the experiments of the prob- 
lems of the preceding article, the orifice was at the center of a 
cylindrical drum 2 feet in diameter, and the plate in which the 
opening was made formed a part of a plane surface closing one 
end of this drum. If an orifice is placed in a relatively small 
pipe, or if there are any obstructions which prevent the liquid 
flowing in freely from all sides, the coefficient will be increased 
and the coefficient of discharge will be greater than 0.62. This 
may be an advantage when it is desired merely to increase the 
flow, but will not do when it is necessary to measure this flow, 
on account of the uncertainty as to the value of the coefficient 
of discharge. 

Fig. 11 shows an orifice in a thin plate. The solid lines B B 
indicate the form of the jet when there is no obstruction to the 
lateral flow behind the plate. The broken lines C C show the 
form of the jet when a hollow cylinder, open at the ends, is 



22 



placed in the liquid behind the plates. -The effect of this cylinder 
is to prevent some of the flow from the sides and to cause the 
liquid to move toward the orifice more nearly normal to its 
plane. This is said to partly suppress the contraction. 

If the cylinder were replaced by a hollow conical frustrum 
with its smaller end, equal to the orifice, in contact with the plate, 
the contraction would be more perfectly suppressed and the co- 
efficient of discharge still further increased. 

If a bell-mouth opening were used, with the small end cyl- 
indrical and equal to the orifice, the change in direction of the 




Fig. 11. 



liquid motion would be gradual, so that there would be no eddy 
losses, the coefficient of contraction would be unity and the 
coefficient of discharge only a little less than unity. 

13. Nozzles. — When it is desired to increase the co- 
efficient of discharge without seriously impairing the velocity of 
the jet, orifices are replaced by nozzles. Fig. 12, I, shows a 
nozzle in the form of a frustrum of a cone ending with a short 
cylinder. There is some little contraction when the liquid enters 
the cone and again when it passes from the cone to the cylinder. 
This causes eddying. The liquid entirely fills the nozzle so that 



2 3 

the coefficient of contraction is nearly unity. The velocity at the 
center is approximately the theoretical velocity, but it is less near 
the surface, so that the average velocity is in the neighborhood 
of 0.95 of the theoretical velocity (depending, of course, on the 
angle of the cone). 

Fig. 12, II, shows a nozzle in which the cone is replaced by 
a surface which gradually enlarges and joins the cylinder on a 




Fig. 12. 




tangent instead of at an angle. The coefficient of contraction 
is unity. There is little loss of velocity due to eddying, if the 
surfaces are properly designed. There is some loss from sur- 
face friction, making the coefficient of velocity about 0.99. 

PROBLEMS 

1. A nozzle 2i inches long was made in the form of a frustrum 
of a cone l-J-^- inches in diameter at the inner end, and f inch in diameter 
at the outer end. Under a head of 6.08 feet it discharged 17.52 cubic 
feet of water in 20 minutes. Find the coefficient of discharge. 



Ans. 



K d = 0.963. 



2. A nozzle in the form of Fig. 12, I, tapered from 1 inch 
diameter to |.|- inch diameter in a length of 2 inches. The cylindrical 
part was |3 inch in diameter and 1 inch long. It discharged 18.18 
cubic feet of water in 25 minutes under a head of 5.41 feet. Find the 
coefficient of discharge. 

Ans. 0.949. 



14. Large Orifice under Low Head. — The velocity at 
the center of a vertical orifice is not exactly the average velocity, 



2 4 



since the velocity varies as the square root of the depth, instead 
of directly as the depth. When the vertical dimension of the 
orifice is small compared with h, the error is inappreciable, but 
when a large orifice is used under relatively small head, allow- 
ance must be made for the variation in velocity. 




Y/////////A 




Fig. 13. 

Fig. 13 shows a rectangular orifice of breadth b, the top of 
which is at a distance h and the bottom at a distance H below 
the surface of the liquid. At a distance y below the surface of 
the liquid, V = V 2 gy, and, if the coefficient of contraction 
were unity, the quantity which would flow through an element 
of area b dy in unit time would be 



X 

dQ = b V~2~g j y 2 dy, 



^ 2 b , 



H 



[y 2 J * 



(1) 
(2) 



Substituting the limits and multiplying by the coefficient of con- 
traction K 



Q 



Kb(H 2 — h 2 ) 



(3) 



EXAMPLE 

A rectangular orifice of width b has the lower edge 0.64 feet and the 
upper edge 0.36 feet below the surface of the water. Find the quantity 



25 



by means of equation (3). Also find the quantity by means of the orifice 
formula using h = 0.50 feet. 

Q=.| K bVT^ (0.512 — 0.216) =0.1973 K b V 2~J] 
= 0.28 K b VI^X V 0750 = 0.1980 K b V2~g. 

It is evident from this example that little error is made by 
taking the velocity at the middle of the orifice as the average 
velocity, even when the head is very low compared with the 
height of the orifice. 

15. Rectangular Weir. — Fig. 14 shows a rectangular 
weir, or notch. The horizontal edge C D, over which the water 
flows, is called the crest. The length of the crest or breadth of 
the weir is b. The depth of the crest below the plane of the water 




RlifSiirfPiIiBT 



HEBBHI 



Fig. 14. 

surface back of the weir is represented by H. This is called the 
depth or height above the crest. Equation (3) of the preceding 
article may be used to find the quantity. With H for the upper 
limit and for the lower limit, (y is taken as positive down- 
ward.) 



0= — Kb\/2gH I = ~KbH\/2gH 
3 3 



Q=—KA VzgH, 



(1) 
Formula III 



where A = b H = the area of the weir below the plane of the 
water surface behind it. 

Formula III differs from Formula II by the coefficient 2 / 3 
and the fact that H is measured from the crest of the weir 



26 



(equivalent to the bottom of the orifice) while h in Formula II 
is measured from the middle of the orifice. 

2 

The product of — K V 2 g is frequently given as a single 
o 
expression, when K = o.62 this product is nearly 3.33 and 
Formula III may be written 

= 3-33^. (2) 

In this form it is called the Francis formula. 

In a rectangular weir there is a contraction at the top and 
bottom and also at each side. If the breadth is increased while 
the head remains constant, the contraction at the top and bottom 
will have the same actual value as before and the same relative. 
The contraction at the sides will have the same actual value but 
the relative value will be less. It follows that, for a given head, 
the coefficient of discharge increases as the width is increased, 
and, for a given width, the coefficient of discharge decreases as 
the head is increased. For a given ratio of breadth to height the 
coefficient of discharge is fairly constant, except for very low 
heads of two or three inches for which the coefficients are larger. 
Table II gives the coefficient of discharge for various ratios of 
breadth to depth. It is derived from the tables of Hamilton 
Smith (see Merriman's Hydraulics). 

TABLE II 

Coefficient of Discharge for Rectangular Weirs with Full 
Contraction 

b K 



H 




10 


0.62 


5 


0.61 


3 


0.60 


2 


0.59 



In order that the coefficient of Formula III may have a 
definite value it is necessary that the weir shall have a sharp edge, 
that the upper surface shall be plane, and that the liquid, after 
passing the edge, shall spring clear without touching. In order 
to have full contraction at the top and bottom, the height of the 
weir from the crest to the bottom of the plane surface should 



be two or three times as great as the height of the water above 
the crest. In order to have full end contraction, the width of 
the plane surface and of the water back of it should exceed the 
length of the crest by four or five times the maximum H. 

If the crest includes the entire width of the stream back of 
the weir, it is called a suppressed weir, and it has a coefficient 
somewhat larger than those given in Table II. 

PROBLEMS 

1. A rectangular weir, 6 inches wide, discharged 25.19 cubic feet 
of water in 10 minutes under a head of 0.0865 feet. Find the coefficient 
of discharge. 

Ans. K = 0.612. 

2. The weir of Problem 1 discharged 1389 pounds of water at a 
temperature of 76 degrees Fahrenheit in 15 minutes under a head of 
0.0788 feet. Find the coefficient of discharge. 

3. The water of Problem 2 flowed from a l T U-inch orifice under 
a head of 18.16 inches. Find the coefficient of discharge of this orifice. 

4. How much water will flow per minute over a rectangular weir 
20 feet wide under a head of 2 feet? How much must the head be 
increased to double the quanitity? 

16. Triangular Weir. — A triangular weir, called also a 
V notch, is shown in Fig. 15. Through an element of area dA 
at a distance y below the surface .of the water back of the weir. 




28 



dQ = yzgy dA. (i) 

dA =■ x dy, 

where x is the length of the element. From similar triangles, 

x H — y by 

— ;*-(* ) (2) 

b H H 



c a r 

Q = 6j V2<7 (3' ! ) dy (3) 



_3 5 ^J" 15 



V2<7 H^ ( 4 ) 
Multiplying by the coefficient of discharge, 



Q = ^-KbH\/ 2 gH. Formula IV. 

Since b = 2 H tan 0, where is the half angle of the notch, 



Q = —K tan V 2 # # 2 . (6) 

The quantity discharged by an orifice varies as the square root 
root of the head. The quantity discharged by a rectangular weir 

3 

varies as H 2 , while the quantity discharged by triangular weir 

5 

varies as H 2 . A four-fold increase of the head doubles the flow 
from an orifice, makes it eight times as great from a rectangular 
weir, and thirty-two times as great from a triangular weir. 

The coefficient of discharge of a triangular weir depends 
upon the angle at the bottom. It is about o.6o. 

PROBLEMS 

1. Find the discharge per second from a right-angled triangular 
notch under a head of 2 feet, if i£ = 0.60. 

Ans. 14.52 cubic feet per second. 

2. Find the discharge of the weir of Problem 1 for a head of 6 
inches. 

3. Find the discharge from a 60 degree triangular weir under a 
head of 8 inches, if the coefficient of discharge is 0.60. 



29 

4. A triangular weir, 5.03 inches wide and 4.06 inches high, dis- 
charged 15.23 cubic feet in 20 minutes under a head of 0.1728 feet. Find 
the coefficient of discharge. 

Ans. i^ = 0.593. 

i/. Velocity of Approach. — In the use of a weir, the 
gage to measure the head is set some distance back of the weir 
where the water has little velocity. When the channel is rela- 
tively large compared with the section b H, the velocity at the 
gage is so small that no correction is necessary; but with a nar- 
row or shallow channel, correction must be made for the velocity 
of approach. A similar correction is necessary when a relatively 
large orifice is used at the end of a pipe. 

The kinetic energy of one pound moving with a velocity of 

V 2 
V feet per second is . This is called the velocity head, and 

must be added to the pressure or height as measured to get the 
effective head. 

EXAMPLE 

Water flows from, an orifice in a thin plate at the end of a pipe. 
The velocity in the pipe is 6 feet per second and the pressure back of 
the orifice is 12 feet of water. Find the velocity in the contracted 
vein. 

36 

Effective head = 12 -) = 12.56 feet of water. 

64.32 



V = V64.32 X 12.56 = 28.42 feet per second. 

PROBLEMS 

1. Water flows over a rectangular weir 5 feet wide. The measured 
head is 8 inches and the velocity of the water at the gage is 2 feet per 
second. Find the quantity if K = 0.61. 

2. A nozzle is placed at the end of a 4-inch pipe. The jet from 
the nozzle is 2 inches in diameter. What is the velocity of the jet when 
the measured head in the pipe is 16 feet of water? 

Ans. 33.15 feet per second. 

18. Relation of Area to Velocity. — Fig. 16 shows a pipe 
which converges and diverges gradually. At one section, the 
area is A x and the velocity is V x . At another section the area is 



30 




Fig. 16. 

A 2 and the velocity is V 2 . Since the same quantity flows past 
both sections in the same time, 

V X A X -V % A V (i) 

Since the area of a circular section varies as the square of the 
diameter, 

V X D\-V % D\; (2) 



V, = v 



>m 



(3) 



PROBLEMS 

1. A 4-inch pipe enlarges to a 6-inch pipe. The velocity in the 
4-inch pipe is 18 feet per second. Find the velocity in the 6-inch pipe. 

2. A 6-inch pipe enlarges to a 10-inch pipe and then contracts to 
an 8-inch pipe. If the velocity in the 6-inch pipe is 16 feet per second, 
find the velocity in the others. 

19. Relation of Velocity to Pressure. — In Fig. 16, the 
pressure at A x is h x , and the potential energy of one pound of 
water is h x foot-pounds. The kinetic energy of one pound of 

water flowing with a velocity V x is , and the total energy 



per pound of water is given by 

Energy = h x + 



V\ 

*g 



(1) 



3i 

In equation (i) the potential energy is calculated with re- 
ference to the axis of the pipe, which is the datum line from 
which h x is measured. If the loss due to friction and eddying 
be neglected, the energy at A 2 is equal to the energy at A lf and 

V\ V\ 
h x -\ = h 2 -\ . Formula V. 

2 9 2 9 

Formula V is called Bernoulli's theorem. 

VI — VI 
K — h 2 = . (3) 

2 9 
The pressures h x and h 2 must be measured from the same hori- 
zontal line. The terms h x and h 2 are called the pressure heads. 



V\ V 



The terms and are called velocity heads. When the 

2 9 29 

pipe is not horizontal a third term is added to give the distance 
from some datum line to the center of the pipe; if the pressures 
are all referred to a single level, this term is unnecessary. 

PROBLEMS 

1. A 12-inch pipe contracts gradually to a 6-inch pipe. The 
velocity in the 12-inch pipe is 5 feet per second and the pressure is 
12 feet of water. Find the pressure in the 6-inch pipe, if both pipes are 
horizontal. 

400 — 25 

Ans. fe = 12 = 6.17 feet. 

64.32 

2. A horizontal pipe contracts gradually from 6 inches to 4 inches 
diameter, and then enlarges gradually to 8 inches. The velocity in the 
6-inch pipe is 12 feet per second and the pressure is 6 feet of water. 
Neglecting the loss, find the pressure in each of the others. 

Ans. — 3.09 feet in the 4-inch pipe ; 7.10 feet in the 8-inch pipe. 

3. An 8-inch horizontal pipe contracts gradually to 4 inches. The 
pressure in the 8-inch pipe is 10 pounds per square inch, and the velocity 
is 8 feet per second. Find the pressure in the 4-inch pipe. 

4. A 6-inch pipe gradually converges to 3 inches diameter. The 
pressure in the 6-inch pipe is 16 feet of water and the pressure in the 
3-inch pipe is 4 feet of water. Find the velocity in each pipe and the 
discharge per second. 

Ans. 7.17 feet per second; 28.69 feet per second; 
2.82 cubic feet per second. 



32 



20. The Venturi Meter. — When a pipe converges 
gradually and then diverges gradually, there is little loss of 
energy, so that an arrangement of this kind may be inserted in 
a pipe line and employed to measure the velocity. If the 
diameter of the pipe at two sections is known, the difference in 
pressure at these two sections, substituted in equation (3) of the 
preceding article, gives the velocity in the pipe. This arrange- 
ment is called a Venturi meter (Fig. 17). The smallest sec- 




Fig. IT. 



tion of the pipe is called the throat. In Fig. 17, a mercury 
column in a U-tube is used to measure the pressure difference. 
Two separate gages of any kind may be employed. As it fre- 
quently happens that the pressure in the throat is. below the 
atmosphere, a separate gage at this point must be one which will 
read a vacuum. 

PROBLEMS 

1. The diameter of the throat of a Venturi meter is one-third 
the diameter of the pipe. The pressure difference is 30 feet of water. 
Find the velocity in the pipe. 

Ans. 15.53 feet per second. 

2. The diameter of the throat is one-half the diameter of the 
pipe, and the pressure difference is 24 inches of mercury. Calculate the 
velocity in the pipe. 

Ans. 10.80 feet per second. 

3. The throat of a Venturi meter is 4 inches in diameter and the 
pipe is 12 inches in diameter. The pressure in the pipe is 20 pounds 



33 



per square inch and the vacuum in the throat is 15 inches of mercury. 
Find the velocity and the discharge in cubic feet per minute. 

4. If n is the ratio of the diameter of the pipe to the diameter 
of the throat, and h is the pressure difference in feet of water, show 
that the -velocity in the pipe is given by 



Vi = 



2 g h 



V n 4 — 1 



5. Find the expression for Vi in terms of h when n =3. 

21. Diverging Mouthpiece. — If a pipe enlarges grad- 
ually so that there is little eddy loss, the pressure is increased. 
If the larger pipe opens into the air, its pressure at the end is 
atmospheric, while that in the smaller pipe is below the atmos- 
phere. 

EXAMPLE 

A i-inch pipe enlarges gradually to one inch in diameter, and dis- 
charges into the air. The velocity in the smaller pipe is 40 feet per 
second. Find its pressure. 

Representing the pressure in the smaller pipe by h 2 , 

40 2 10 2 

h 2 + — _ = + - -; 



fk 



64.32 64.32 

-23.32 feet of water. 



Fig. 18, I, shows a bell-mouth nozzle. The coefficient of 
contraction is unity and the velocity is \/ 2 g h, where h is 





Fig. 18. 



34 

measured from the top of the water to the center of the orifice. 
Fig. 1 8, II, shows a diverging extension added to this nozzle. 
The pressure at the throat is below atmospheric, so that the ef- 
fective pressure to produce velocity in the throat is the sum of 
the head and this vacuum. The velocity in the throat is, there- 
fore, greater than V 2 9 h, and the coefficient of discharge, in 
terms of the area of the throat, is greater than unity. The 
velocity diminishes in the diverging extension and the final 
velocity into the air is, of course, a little less than V 2 g h. 

EXAMPLE 

A nozzle contracts gradually to a diameter of one inch in the 
throat and then enlarges to a diameter of 1.5 inches. If the length of 
the extension is such that the enlarged portion runs full, and if there 
is no loss of energy, find the velocity in the throat under a head of 4 
feet of water. 

The velocity of discharge into the air with no loss of energy is 
given by 



F 3 = V 64.32X4 = 16.03 feet per second. 

The velocity in the throat will be 16.04 X 2:2:5 = 36.09 feet per second. 
The effective head to produce a velocity of 36.09 feet per second is 
20.25 feet of water, and the vacuum at the throat is 20.25 — 4 = 16.25 
feet of water. The vacuum in the throat might have been calculated by 
Bernoulli's theorem. The coefficient of discharge in terms of the area 
of the throat would be 2.25 if there were no loss of energy. 

If there is too great expansion, or if there is a large pres- 
sure giving a high velocity in the throat, the vacuum may become 
so low that the water will boil. The velocity will then no longer 
be inversely proportional to the area of the section (since at the 
throat the volume will be increased by the volume of the steam) 
and Bernoulli's theorem will not hold good. . 

PROBLEMS 

1. A nozzle converged to a diameter of -j-|. inch and then enlarged 
gradually to a diameter of one inch. It discharged 24.62 cubic feet 
of water in ten minutes under a head of 0.9617 foot. Find the coeffici- 
ent of discharge in terms of area of the throat, using the equation 
Q = KAV2gh. Ans. if = 1.45. 

2. In Problem 1, find the coefficient of discharge in terms of the 
area of the opening at the end. Ans. K = 0.96. 

3. In Problem 1, what was the probable vacuum at the throat? 



35 

Since there is a partial vacuum in the throat of a diverging 
mouth-piece which opens into the air, it may be used to lift water 
from a lower level. Fig. 19 shows a diverging mouth-piece to 
which a vertical pipe B is connected at the section at which the 



V. n'n M^SSSSSNSS 




Fig. 19. 

divergence begins. If the distance h from the surface of the 
water below the pipe is less than the vacuum at the throat, some 
of the water will be forced up through B and increase the quan- 
tity passing through the section C. 



EXAMPLE 

In Fig. 19, the area at the throat is 4 square inches and the area 
at C is 10 square inches. The area of B is 2 square inches. The 
velocity in the throat is 20 feet per second and the velocity in B is 5 
feet per second. Find the vacuum at the throat and the distance the 
water may be lifted, if there is no loss by friction. 

20X4 + 5X2 

Velocity at C — = 9 feet per second. 

10 



36 



400 — 81 

Vacuums = 4.95 feet. 

64.32 

To give the water in B a velocity of 5 feet per second requires 

25 



0.39 foot of water. 



64.32 

/* = 4. 95 — 0.39 = 4.56 feet. 

22. Loss of Energy at Abrupt Enlargement. — When a 
pipe enlarges suddenly, as in Fig. 20, there is a loss of energy. 
The water in the larger pipe may be considered as a large mass 
moving with uniform velocity V. 2 and the water flowing from 




Fig. 20. 

the smaller pipe may be regarded as a stream of small particles 
coming into collision with this large mass and suffering a change 
of velocity from V ^ to V 2 . The problem is that of a stream of 
inelastic bodies striking a large moving body. 

When a body of mass m changes its velocity from V 1 to V 2 , 

m{V\-V\) 



it gives up energy amounting to 



2 9 



foot-pounds. 



Part of this energy is lost in the form of heat and sound vibra- 



37 







"z 




m h 


M 


I 


v, 




X- 


H 






y 2 


M 




to 





Fig. 21. 
moves during collision 



tions and part of it does work on the 
second body. Fig. 21, I, shows a 
small mass m overtaking a large mass 
M. Fig. 21, II, shows the two bodies 
at the end of the collision, when 
the velocity of m is reduced to V 2 . 
If t is the time of collision in 
which the velocity changes from 
V x to V 2 , the distance which M 
is given by x = V 2 t. The negative 



acceleration of m is 



V 1 — V 2 
t 



, and the average force of m on M 



is 



m(V 1 --V 2 ) 



pounds. The work during collision, being the 



force times the displacement, is given by 

»t ( Vi — V 2 ) VJ m ( V x — V 2 ) V, 



Work == 
m{V\ — V\) 

The expression 



gt 

m(V x 



Vo)V, 



g 

m (V, 



m ( V 



V 2 Y 



2g 



(I) 

(2) 

-* y 

represents the lost energy. 



v t y 



In the case of a sudden change of liquid velocity, this energy 
is partly converted into heat and partly expended in giving 
rotary motion to the liquid in the eddies. Bernoulli's theorem 
for sudden enlargement is 



*! + 



v\ 



K + 



vi 



{V. — V.Y 



(3) 



2CJ 2CJ 2CJ 

At a sudden enlargement there is an increase of pressure. 
but this increase is less than in the case of a gradual enlarge- 
ment on account of the lost energy represented by the last term 
of equation (3). 

PROBLEMS 

1. A 4-inch pipe enlarges suddenly to 8 inches. The velocity in the 
4-inch pipe is 20 feet per second and the pressure is 12 feet of water. 
Find the pressure in the 8-inch pipe. 

Ans. 2.33 feet. 

2. A 2-inch pipe enlarges gradually to 4 inches diameter and then 
converges gradually to 2 inches. Find the difference in pressure in the 
2-inch pipes if the velocity in the 2-inch pipes is 40 feet per second. 

Ans. 14.0 feet. 



38 



23. Loss at Abrupt Entrance. — Fig. 22 shows an abrupt 
entrance into a pipe. There is a contracted vein, such as occurs 

when a liquid jet flows from an 
orifice in a thin plate. The space 
in the pipe surrounding the jet 
is filled with eddying liquid. 
When the stream enlarged from 
the contracted vein to the full 
section of the pipe, there is a 
loss of energy amounting to 

(v 1 — vy A f ' .. 

— per pound 01 liquid, 

where V is the velocity in the 
pipe, and V 1 is the velocity in 
the contracted vein. With the 
coefficient of contraction, 0.62, 
V= 0.62 V x , hence, ^=1.61 V 
and 




Fig. 22. 



Loss 



(1.61V— vy 0.37V 2 



(I) 



2 9 zg 

This is called the entrance head, or the loss of head at entrance. 



PROBLEM 

Water flows through a pipe with a velocity of 12 feet per second. 
If the entrance is abrupt, how many feet of water are required for 
the lost head? 

24. Standard Short Pipe. — A standard short pipe is a 
short nozzle with an abrupt entrance, as in Fig. 22. The length 
is about two and one-half times the diameter. Since the energy 

0.37 v ? 



lost at entrance is 



2 9 

1 • 37 V 2 = 2 g h 
V 



the velocity is given by 



(1) 
(2) 



0.85 V 2gh. 

Equation (2) would give the coefficient of velocity if the velocity 
were constant across the entire section. In reality, the velocity 
at the center is \J~2 g h, while that at the surface is only about 
two-thirds as great. The average velocity is found to be 



39 

o.S2\/2gh. The coefficient of contraction is unity, so that the 
coefficient of discharge is about 0.82. 

Since the water in the center is moving much faster than 
at the circumference of the jet, the stream quickly breaks into 
drops at the surface and has an appearance quite different from 
that of a jet flowing from an orifice in a thin plate. 

If the flow is started carefully without wetting the inside 
surface of the pipe, the jet may spring clear, and the pipe will 
behave like an orifice in a thin plate. If then the outer end of 
the pipe is closed for an instant, it will change the character of 
the jet. The coefficient of contraction will change from 0.62 
to 1.0 and the coefficient of velocity from 1.0 to 0.82. The 
range of the jet will be diminished and the quantity increased. 

PROBLEM 

1. A short pipe, 1 inch inside diameter, and 2J-1- inches long, dis- 
charged 50.5 cubic feet of water in 578 seconds under a head of 5.97 
feet. Find the coefficient of discharge. 

Ans. # = 0.818. 

25. Liquid Friction. — Experiments with the friction of 
liquids on solids give the following : 

(a) The friction between a solid and a liquid varies as the 
square of the velocity. (The friction of a solid on solid is in- 
dependent of the velocity.) 

(b) The friction between a solid and a liquid varies as 
the area of the surface of contact. (The friction of solid on 
solid is independent of the area of contact, unless the area is so 
small that the bearing pressure exceeds the elastic limit.) 

(c) The friction between a. solid and a liquid is inde- 
pendent of the pressure. (The friction of solid on solid varies 
directly as the pressure.) 

The force required to move a surface of area A through a 
liquid with a velocity of V feet per second is K A V 2 pounds, 
where K is the force required to move one square foot of the 
surface with a velocity of one foot per second. The motion is 
parallel to the surface, so that the liquid flow with reference 
to the surface is tangent to the surface. The same force is re- 
quired if the surface is stationary and the liquid is moved 
tangent to it. 



40 



In a pipe of diameter D feet and length L feet, the surface 
is D L and the force in pounds required to keep up a velocity of 
V feet per second is K D L V 2 . If the pressure difference of 
the ends of the pipe is h feet of water, the pressure difference 
in pounds of the ends of the cylinder of the liquid which fills 



the pipe is 
liquid. 



wD 2 h 



where w is the weight of a cubic foot of the 



wD 2 h 



= KuD L V 2 



(i) 



4KLV 2 

D w 



(2) 



The term 
becomes 



4 K 



w 



may be replaced by and equation (2) 



zg 



kL V'< 



h = 



D 2<7 
The equation is put in this form with 



(3) 



V 2 

zg 



as a term in 



order that it may be conveniently used along with the entrance 
head and velocity head in a pipe. The constant k depends upon 
the nature of the surface and the material of which it is made. 
It also depends somewhat on the diameter, but for the present 
this term will be neglected. The head h of equation (3) is called 
the friction head. 

PROBLEMS 

1. What is the friction head required to force water with a 
velocity of 8 feet per second through a 6-inch pipe 1200 feet long, if 
K = 0.03. 

Ans. ^ = 71.6 feet. 

2. In a test of a 1-inch pipe the difference in head between two 
sections 11 feet 4 inches apart was 1.66 feet when the velocity was 5.2 
feet per second. Find k. 

Ans. k = 0.029. 

3. In the pipe of Problem 1, the mean difference of head was 2.42 
feet and the discharge was 2000 pounds of water at a temperature of 
76 degrees Fahrenheit in 15 minutes and 8 seconds. Find k from this 
test. 



41 

The head required to give a velocity of V feet per second 

V 2 
is , and the lost head at entrance, when the entrance is 

O X7 V 2 

abrupt, is — — , so that the entire head required to force 

water into and through a pipe of length L and diameter D is 
given by 

V 2 kL 

Total head = ( i + 0.37 -\ ) . Formula VI. 

2g D 

The first term of Formula VI is the velocity head; the sec- 
ond term is the entrance head; the third term is the friction 
head. If the entrance is gradual, the entrance head vanishes, 
leaving only two terms. 

PROBLEMS 

4. What is the total pressure in a tank to force water into and 
through a 4-inch pipe 50 feet long with a velocity of 12 feet per second, 
if the entrance is abrupt and £ = 0.03? 

Ans. 13.1 feet. 

5. Solve Problem (4) for a velocity of 20 feet per second. 

Ans. Total head = 6.22 (1+ 0.37 + 4.5) = 36.51 feet. 

6. A 6-inch pipe 200 feet long is connected to a tank with abrupt 
entrance. The friction coefficient of the pipe is 0.03, and the outlet of 
the pipe is 80 feet below the top of the water in the tank. Find the 
velocity in the pipe and find what part of the head is used to overcome 
friction and what part is used for velocity and entrance head. 

Ans. V = 19.38 feet per second. 

Fig. 23 shows the pressure drop in Problem (5). The 
friction head is 27.99 ^ eet an d the entrance and velocity heads 
are together 8.52 feet. There is a sudden drop of pressure at 
the entrance, of which 2.30 feet represent lost energy, and 6.22 
feet represent the kinetic energy of the moving liquid. There 
is a uniform drop in pressure for the length of the pipe, indi- 
cated by the broken line. This is called hydraulic gradient. It 
represents the friction head. When the liquid issues from the 
pipe it has no pressure head, but it still retains the kinetic 
energy equivalent to a head of 6.22 feet. If the pipe ended in 
a second tank, and the entrance were gradual, so that there 



42 




was no loss of energy, the water would flow with a velocity of 
20 feet per second when the water level in the second tank was 
nearly 6.22 feet above the end of the pipe. 

PROBLEMS 

7. Water flows from a tank into a horizontal pipe 500 feet long 
and 4 inches in diameter. The entrance is gradual so that there is no 
entrance loss. If k is 0.03 and the head in the tank is 100 feet, find 
the velocity and draw the hydraulic gradient. 

8. Water flows from a tank into a 6-inch pipe 100 feet long. 
From the 6-inch pipe it flows into a 4-inch pipe 50 feet long. The 
entrance into the 6-inch pipe is sudden and into the 4-inch pipe is 
gradual. The velocity in the 6-inch pipe is 8 feet per second. Draw 
the hydraulic gradient if k = 0.03. 



26. Pitot Tube. — When a tube, open at the end, is held 
in a water jet with the plane of the opening normal to the di- 
rection of motion of the liquid, a pressure is developed in the 

v- 

tube equal to the head . Fig. 24 shows a tube of this kind. 

The end of the tube is parallel to the jet, and is cut off square so 
that the moving liquid strikes normal to the section. The tube 
is curved upward and the water in it rises to a height equal to 
that of the water in the tank, provided the water flows from an 



43 




Fig. 24. 



orifice in a thin plate with a velocity 
coefficient of unity. This arrange- 
ment is called a Pitot tube. 

A Pitot tube may be used to find 

the velocity in a pipe. Its reading 

is the sum of the velocity head and 

the pressure head. To find the 

velocity it is necessary to make a 

separate pressure reading. Fig. 25 

shows a Pitot tube placed in a pipe 

with the pressure tube connected at 

the top of the pipe in the section at 

the end of the Pitot. Frequently the 

pressure tube and the Pitot tube are 

constructed together as one piece. In 

that case the apparatus is so made that the water flows tangent 

to the end of the pressure tube and strikes the wall of the Pitot 

tube normal. 

PROBLEMS 

1. Water flows from an orifice in a thin plate. A Pitot placed in 
the jet reads 18 inches of mercury. Find the velocity of the water. 

Ans. 36.22 feet per second. 

2. A Pitot and a pressure tube are placed in a pipe. The Pitot 
reads 20 pounds per square inch and the pressure tube reads 30 inches 
of mercury. Find the velocity in the pipe. 

3. A small Pitot was placed in a jet from a 1-inch short pipe. At 
the middle of the jet the Pitot reading was 5.76 feet when the pressure 
in the tank to which the pipe was connected was 5.76 feet. At 0.3 inch 
from the center of the pipe the Pitot read 4.63 feet when the tank 
pressure was 5.70 feet. At the surface of the jet the Pitot read 2.35 
feet when the tank pressure was 5.18 feet. Find the ratio of the 
velocity at each of these places to the theoretical velocity. 



Instead of moving water striking a stationary tube we may 
have the tube moving and the water stationary. The pressure 
developed in both cases is the same. This is the principle used 
in scooping up water into the tender of a moving train. 

Since water striking against a tube at any angle produces 
some pressure, it is necessary that a tube for measuring the 
static pressure of moving water should be exactly normal to the 



44 



pipe. In Fig. 25, tube B gives the correct pressure. Tube A 
gives a pressure which is too low and tube C gives a pressure 

which is too high. 



^ 



\h 



ABC 



\\\\\\\\\\\\\\\\\\\\\\\\\\ 




ys^ssys^sssssssssssssss^ysssssssy 



^V.W'ASSSY.Y.'iSW 



Fig. 25. 



When a Pitot tube is used to measure velocity in a pipe, 
the tube itself forms some obstruction and consequently reduces 
the area of the section and increases the velocity above that of 
the rest of the pipe, so that, when a large Pitot is used in a rela- 
tively small pipe, there may be considerable error in the pres- 
sure reading. 



■SSL.* C0N( *ESS 



0020588 785 4 '% 



